# Converting from Rectangular Coordinates to Polar - Concept

###### Explanation

We will often be asked to convert rectangular to polar coordinates, and this conversion will be very important to understand in Calculus. In order to **convert rectangular to polar coordinates**, we use the distance formula to find the radius, and the inverse tangent function to find the angle. We may also sometimes be asked to convert from polar coordinates to rectangular coordinates.

###### Transcript

Converting from rectangular coordinates to polar coordinates. And that can be kind of tricky because remember that the polar coordinates for a point are not unique. So let's make a rule here that we're going to get r to be greater than or equal to 0 and theta between 0 and 2 pi. This will allow us to get a unique set of polar coordinates for a point and there will be a lot less confusion. Your teacher may actually have a requirement like that as well.

So all of these are rectangular coordinates. I want to convert them to polar. So let's start with this one. So x is root 3 and y is 3. Now the first thing I can do is use this formula here to find the r value, right? The distance of the point from the pole. So r squared is root 3 squared plus 3 squared. That's 3+9 12. So r=2 root 3. Remember you should the positive value for r so there's no plus or minus here. We want the positive value. And then we can use these formulas to find theta. So you have x=r cosine theta and that means cosine theta equals x over r. Now x is root 3, r is 2 root 3. And so that's one half. And I also use this formula, right? y=r sine theta. So sine theta is y over r. The y value is 3, the r value is 2 root 3 and I can rationalize this denominator by multiplying by root 3 over root 3 and I get 3 root 3 over 2 times 3. The threes cancel. Root 3 over 2. So the question is what angle has a cosine of one half and a sine of root 3 over 2?

Well, that's pi over 3. Theta's pi over 3. And remeber we want theta to be between 0 and 2 pi. I mean there is another angle of 7 pi over 3 that works but it's not in the interval that we want. So this is the theta that we want and the polar coordinates are going to be, remember r comes first, so 2 root 3 theta. Pi over 3.

Okay, let's try another example. The point -5 5. First, r squared equals x squared plus y squared. So -5 squared plus 5 squared. That's 25+25 or 50. And that means that r equals 5 root 2. Again, we picked up the positive value. And we're going to use the fact that cosine theta is x over r, -5 over 5 root 2 and that's -1 over root 2 which is the same as -2 over root 2 over 2. And sine theta is y over r. So it's 5 over 5 root 2, 1 over root 2 which is root 2 over 2. Now what angle has a cosine that's negative root 2 over 2 and a sine that's positive root 2 over 2? You could draw a little unit circle if you like. We're over here, right? A negative a negative cosine value and a positive sine value. This is going to be a reference angle of 45 degrees. The angle's 135 which in radiance is 3 pi over 4. Now that means my point is going to be r first 5 root 2 theta, 3 pi over 4.

And finally let's do 0 -10. Actually this one's pretty easy. If you if you graph this point, you can kind of see that well, we can't really use an angle of negative pi over 2 because it's not in the interval that we want. But you can use an angle of 3 pi over 2. So theta would be 3 pi over 2. And remember that r is just a distance from the origin, which is 10. So this would be 10 3 pi over 2. So it's much easier if you're converting a point that is on a coordinate axis in the polar coordinates.